A pendulum swings back and forth with a period of 0.5 s. What is the length of the pendulum arm?

Question

i know the answer is 0.06 but which formula do i use?

Answer 1

Best Answer - Chosen by Asker

L = g/(4*pi2*f2)
g=32ft/(s^2)
f=(60/t)/60, or 1/0.5s = 2.0 Hz

Solve for L,
L=[32ft/(s^2)]/[4*(pi^2)*(2^2)...
[32 ft/(s^2)]/[16*(pi^2)]
32/[16*9.870]=32/157.9136704 = approx. 0.20 ft or 2.43 inches

In Metric terms
g=9.8m/s^2
the rest of the calculation remains the same, so
9.8/157.9136704 = .0621 m = 6.21 cm

I found the answer to your problem on yahoo answers.
http://ph.answers.yahoo.com/question/index?qid=20070130201536AAna7Rp

Answer 2

The formula is

T = 2pi(L/g)^0.5 where:
T is the periodic time
L length of the pendulum arm
g is the gravity

in your case you need L
L=(T/2pi)^2 * g
=(0.5/(2*3.14))^2 * 9.81
=(0.5/6.28)^2 * 9.81
=0.006 * 9.81
=0.062 m
=6.2 cm